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3 years ago

A rectangular picture frame measures 57 cm by 39 cm. What is the perimeter or distance around the picture frame in meters?

Mathematics

2 answers:

Tamiku [17]3 years ago

8 0

The answer is 192 cm

Svetlanka [38]3 years ago

4 0

(57*2)+(39*2)=192cm now that we have the perimeter we divide it by 100 because 100cm equal one meter so therefore u should get 1.92 meters and if u want it in fraction form 1 4/25 meters

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olga_2 [115]

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3 years ago

In the diagram AB is parallel to CD. Calculate the value of A ?

olga55 [171]

The image of the diagram is missing, so i have attached it.

Answer:

a = 30°

Step-by-step explanation:

From the diagram attached, we can see that the 2 angles a and 5a are interior supplementary angles.

The sum of interior supplementary angles is 180°

Thus;

5a + a = 180

6a = 180

a = 180/6

a = 30°

4 0

3 years ago

Let the universal set, s, have 66 elements. a and b are subsets of s. set a contains 19 elements and set b contains 34 elements.

Otrada [13]

The cardinality of the union between two sets is given by

$N(A\cup B) = N(A) + N(B) - N(A \cap B)$

In fact, if an element is in both A and B, it is counted in both N(A) and N(B), so we're counting it twice. By subtracting the cardinality of the interserction, we're taking back one of these "extra-counted" element, and the count is correct. In your case, the numbers are

$N(A \cup B) = 19+34-10 = 43$

This means that the two sets cover, without repetitions, 43 of the 66 elements in the sets. So, there are

$66-43 = 23$

elements which are in neither A nor B

5 0

3 years ago

A company that manufactures cell phones has been given a quarterly operating budget of $1,176,912.42. The company's quarterly op

amm1812

Answer:

maximum is 16,853 cell phones

≤ 18.853

Step-by-step explanation:

1,176,912.42. per quarter budget

247,638.00 per quarter fixed cost

We subtract then divide the amount cost of phones and ensure its less or rounded down.

1,176,912.42 -247,638 = 929274.42

929274.42 / 55.14 = 16853

16,852 < m ≤ 16,853

6 0

3 years ago

Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$